Two cells, having the same e.m.f. are connected in series through an external resistance R. Cells have internal resistances r1 and r2 (r1 r2) respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is :-

Question

Two cells, having the same e.m.f. are connected in series through an external resistance R. Cells have internal resistances r1 and r2 (r1 > r2) respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is :- (A) r1 + r2 (B) r1 - r2 (C) (r1 + r2/2) (D) (r1 - r2/2)

Tardigrade Admin · Accepted Answer

Current in the circuit I=(2 E/R+r1+r2) ∴ Potential difference across first cell V = E - I r 1= E -(2 Er 1/( R + r 1+ r 2))=0 On solving we get, R =( r 1- r 2) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/488c63f5c8bc02e8f96972a699bd7b45-.png" />

Q. Two cells, having the same e.m.f. are connected in series through an external resistance $R$ . Cells have internal resistances $r_{1}$ and $r_{2} (r_{1} > r_{2})$ respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of $R$ is :-

$r_{1} + r_{2}$

$r_{1} - r_{2}$

$\frac{r _{1} + r _{2}}{2}$

$\frac{r _{1} - r _{2}}{2}$

Current in the circuit
$I = \frac{2 E}{R + r _{1} + r _{2}}$
$∴$ Potential difference across first cell
$V = E - I_{r_{1}} = E - \frac{2 E r _{1}}{( R + r _{1} + r _{2} )} = 0$
On solving we get, $R = (r_{1} - r_{2})$

Q. Two cells, having the same e.m.f. are connected in series through an external resistance R. Cells have internal resistances r1​ and r2​(r1​>r2​) respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is :-

r1​+r2​

r1​−r2​

2r1​+r2​​

2r1​−r2​​